The area of the region bounded by the lines $\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4, x=0$ and $y=0$ is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $56 \sqrt{3} ab$
$\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4$
area = $\frac{1}{2}×28\sqrt{3}a×4b$
$=56 \sqrt{3} ab$